Contoh Soal 5 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: y=\sqrt{x^{2}-2x+1}+\sqrt{x^{2}+2x+1}\\ &\textrm{maka nilai}\: \: y=....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle \displaystyle 2x&&&&&\textrm{D}.&\color{red}\displaystyle \left | x-1 \right |+\left | x+1 \right |\\ \textrm{B}.&\displaystyle 2(x+1)&&\textrm{C}.&\displaystyle 0&&\textrm{E}.&\displaystyle \textrm{tidak ada yang benar} \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui}\: \: y=\sqrt{x^{2}-2x+1}+\sqrt{x^{2}+2x+1}\\ &\Leftrightarrow y=\sqrt{(x-1)^{2}}+\sqrt{(x+1)^{2}}\\ &\Leftrightarrow y=\color{red}\left | x-1 \right |+\left | x+1 \right | \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: x\: \: \textrm{bilangan real dan}\: \: 4y^{2}+4xy+x+6=0,\\ &\textrm{maka nilai semua nilai}\: \: x\: \: \textrm{agar nilai}\: \: y\: \: \textrm{real adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle x\leq -2\: \: \textrm{atau}\: \: x\geq 3&&&&&\textrm{D}.&\displaystyle -3\leq x\leq 2\\ \textrm{B}.&\displaystyle x\leq 2\: \: \textrm{atau}\: \: x\geq 3&&&\displaystyle &&\textrm{E}.&\displaystyle -2\leq x\leq 3\\ \textrm{C}.&\displaystyle x\leq -3\: \: \textrm{atau}\: \: x\geq 2 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&4y^{2}+(4x)y+(x+6)=0\\ &\textrm{agar}\: \: y\: \: \textrm{real, maka}\: \: D=b^{2}-4ac\geq 0\\ &(4x)^{2}-4(4)(x+6)\geq 0\\ &\Leftrightarrow 16x^{2}-16(x+6)\geq 0\\ &\Leftrightarrow x^{2}-x-6\geq 0\\ &\Leftrightarrow (x-3)(x+2)\geq 0\\ &\Leftrightarrow \color{red}\displaystyle x\leq -2\: \: \textrm{atau}\: \: x\geq 3 \end{aligned}\\\\ &\textrm{Jika kita diletakkan dalam garis bilangan}\\ &\textrm{akan tampak seperti berikut}\\\\ &\begin{aligned}&\underset{\begin{matrix} \color{red}\Downarrow\\  \color{red}\Downarrow \end{matrix} }{\textrm{Daerah Positif}}\\ &\begin{array}{cc|ccc|ccc}\\ &&&&&&\\ +\: +&+\: +&-\: -&-\: -&-\: -&+\: +&+\: +\\\hline &-\displaystyle 2&&&&3&&\textbf{Sumbu-X}\\  \end{array}\\ &\begin{matrix} \textrm{dengan}\\ \textrm{titik}\: \: \: \: \: \: \\ \textrm{uji}\qquad\\ \color{red}x=0\quad \end{matrix}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah negatif}}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah positif}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{x(x-1)}{2}\: ,\: \textrm{nilai}\: \: f(x+2)= ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle f(x)+f(2)&&&&&\textrm{D}.&\displaystyle \frac{xf(x)}{x+2}\\ \textrm{B}.&\displaystyle (x+2)f(x)&&&\displaystyle &&\textrm{E}.&\color{red}\displaystyle \displaystyle \frac{(x+2)f(x+1)}{x}\\ \textrm{C}.&\displaystyle x(x+2)f(x) \end{array}\\\\ &\textbf{Jawab}:\textbf{E}\\ &f(x)=\displaystyle \frac{x(x-1)}{2},\\ &f(x+1)=\displaystyle \frac{(x+1)(x)}{2}\Leftrightarrow \displaystyle \frac{x+1}{2}=\frac{f(x+1)}{x}\: ,\: \textrm{maka}\\ &\begin{aligned}f(x+2)&=\displaystyle \frac{(x+2)((x+2)-1)}{2}=\frac{(x+2)(x+1)}{2}\\ &=\displaystyle \frac{f(x+1)}{x}.(x+2)=\color{red}\displaystyle \displaystyle \frac{(x+2)f(x+1)}{x} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Jika}\: \: r_{1}\: \: \textrm{dan} \: \: r_{2}\: \: \textrm{merupakan dua penyelesaian }\\ &\textrm{dari persamaan}\: \: x^{2}+px+8=0\: ,\: \textrm{maka}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle \left | r_{1}+r_{2} \right |>4\sqrt{2}&&&\textrm{D}.&\displaystyle r_{1}<0\: \: \textrm{dan}\: \: r_{2}<0\\ \textrm{B}.&\displaystyle \left | r_{1} \right |>3\: \: \textrm{dan}\: \: \left | r_{2} \right |>3 &\displaystyle &&\textrm{E}.&\displaystyle -\left | r_{1}+r_{2} \right |<4\sqrt{2}\\ \textrm{C}.&\displaystyle \left | r_{1} \right |>2\: \: \textrm{dan}\: \: \left | r_{2} \right |>2 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui PK}:x^{2}+px+8=0\\ &\textrm{dengan}\left\{\begin{matrix} r_{1}\\  r_{2} \end{matrix}\right.\: \: \textrm{akar-akarnya, maka}\\ &\bullet \quad r_{1}+r_{2}=-p\\ &\bullet \quad r_{1}\times r_{2}=8\\ &\textrm{Asumsikan kedua akarnya real dan berbeda,}\\&\textrm{sehingga kita pilih nilai} \: \: \: D=b^{2}-4ac>0\\  &\Leftrightarrow (-p)^{2}-4.1.8>0\\ &\Leftrightarrow p^{2}-32>0\\ &\Leftrightarrow (\left | r_{1}+r_{2} \right |+\sqrt{32})(\left | r_{1}+r_{2} \right |-\sqrt{32})>0\\ &\Leftrightarrow \left | r_{1}+r_{2} \right |<-\sqrt{32}\: \: \textrm{atau}\: \: \color{red}\left | r_{1}+r_{2} \right |>\sqrt{32}\\&\qquad\overset{\begin{matrix} \Downarrow\\  \Downarrow \end{matrix}}{\textbf{tidak mungkin}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Jika titik}\: \: (1,y_{1})\: \: \textrm{dan}\: \: (-1,y_{2})\: \: \textrm{terletak pada}\\ &\textrm{grafik}\: \: y=ax^{2}+bx+c\: \: \textrm{dan}\: \: y_{1}-y_{2}=-6\\ &\textrm{maka nilai}\: \: b=....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle \displaystyle -3&&&&&\textrm{D}.&\displaystyle \sqrt{ac}\\ \textrm{B}.&\displaystyle 0&&\textrm{C}.&\displaystyle 3&&\textrm{E}.&\displaystyle \displaystyle \frac{a+c}{2} \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui}\: \: y=ax^{2}+bx+c,\: \: \textrm{untuk}\\ &\bullet \quad (1,y_{1})\Rightarrow y_{1}=a+b+c\\ &\bullet \quad (-1,y_{1})\Rightarrow y_{2}=a-b+c\\ &\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ -\\ &\qquad\qquad y_{1}-y_{2}= 2b=-6\Rightarrow b=\color{red}-3\end{array}$.